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October 27

Crude oil production

Hello,

Can anyone tell me what has been the total world production of crude oil, in trillions of tonnes, over the period 1859 to 2007?

Many thanks, Carlyon —Preceding unsigned comment added by Carlyon (talk • contribs) 00:16, 27 October 2007 (UTC)[reply]

Thats a tough one. Using this dataset from OECD, the total world production of crude oil between 1971 to 2005 was 1.107827 trillion tonnes (trillion meaning a million million). This table provides data back to 1960 in million barrels a day. Beyond that all I can find is Image:Hubbert world 2004.png, from which you could try to source the raw data back to 1900. Rockpocket 01:02, 27 October 2007 (UTC)[reply]

Calcium Gluconate Chemical Structure

I'm having trouble putting together the chemical structure of calcium gluconate. I've tried using the BKchem molecular editor or similar programs to lay it out but i've completely forgotten how to do so. If this was all still fresh in my mind from my school years, I'd have no problem. Wikipedia shows a good bit of information about calcium gluconate, but in particular I need to see in diagram form the chemical structure of said supplement. Again, I've completely forgotten how to put these things together. [1] —Preceding unsigned comment added by 76.177.150.63 (talk) 00:39, 27 October 2007 (UTC)[reply]

It's a salt of gluconic acid. The rightmost hydrogen in the structure is to be removed (and a superscripted minus sign is to be written right of the rightmost oxygen) to get gluconate. Calcium gluconate is made up of 2 of these units and 1 Ca²⁺ ion, compare with calcium acetate (I think the picture in this article is of the kind you want to have). I don't know the real spatial arrangement of calcium and gluconate ions which can be determined using X-ray crystallography. Icek 03:00, 27 October 2007 (UTC)[reply]

Let me see if I've got this straight. From the gluconic acid you've shown, remove the rightmost H and replace it with a superscripted -? How would i illustrate 2 of these units correctly? The shown unit enclosed in parentheses (or whatever the correct term would be here)? Then followed with Ca²⁺? Is this the only possible way of illustrating Calcium Gluconate? thank you very much. —Preceding unsigned comment added by 76.177.150.63 (talk) 23:25, 27 October 2007 (UTC)[reply]

You could also draw 2 gluconate molecules and a Ca²⁺ between them (with the right molecule being a rotated version of the left one - the Ca²⁺ should be close to both O^-), similar to the picture in the Wikipedia article (which is a rendering of a space-filling 3d model - the green thing is the calcium ion). Icek 09:08, 28 October 2007 (UTC)[reply]

3 dB per octave?

Is there a way of getting a 3dB per octave boost or cut over a wide frequency range? —Preceding unsigned comment added by 88.109.17.174 (talk) 00:45, 27 October 2007 (UTC)[reply]

Yes but if it's the simple answer I think it is, I suspect this is a homework problem. Hint: Single poles and zeros create 3dB/octave changes in power. --DHeyward 07:03, 27 October 2007 (UTC)[reply]

Uh, no, first-order filters are 6dB per octave.

But yes, it’s certainly possible to design a filter that only changes by 3 dB per octave. Perhaps the easiest way to design such filter is to use a digital filter. Just do an Fast Fourier Transform, manipulate the spectrum to have the desired frequency response, and convert back to the time domain.

It'd be a lot more complicated to design if you have to do it as an analog filter, but it'd still be possible. It'd have to be a high-order filter, that combines an equal number or almost equal number of high-pass and low-pass first-order filters within one circuit. To see that it’s possible to design such an analog filter, you could always split the signal into an arbitrarily large number of bands using a bunch of band pass filters, amplify each band separately by an appropriate amount, and then join the signals together in one output. MrRedact 08:09, 27 October 2007 (UTC)[reply]

Actually, I thought of a much easier way to design such an analog filter. Just combine a first-order filter with an analog multiplier in a feedback circuit such that the multiplier is doing a square root. MrRedact 08:20, 27 October 2007 (UTC)[reply]

Oops, wait, I think DHeyward may have hit the nail on the head. Are you talking about 3dB voltage per octave, or 3dB power per octave? I assumed you meant voltage. If you meant power, then the answer is a trivial first-order low pass filter or high pass filter. MrRedact 10:06, 27 October 2007 (UTC)[reply]

It was 3 dB per octave in voltage i wanted (power is dead easy). —Preceding unsigned comment added by 88.111.67.141 (talk) 15:44, 27 October 2007 (UTC)[reply]

OK, then did you understand my idea for the analog version of the filter? The input of the whole filter would be the input of a first-order filter. The output of the first-order filter would go to the + input of an op amp. The output of the op amp, which is the output of the whole filter, would also go to both inputs of a four quadrant analog multiplier. The output of the multiplier would go to the – input of the op amp. You might need a resistor or two in there to tweak the gain or bias, but that’s the basic idea. MrRedact 20:39, 27 October 2007 (UTC)[reply]

Yes I understand: a 6dB/octave roll off, when square rooted will give a 3dB/oct roll off? This is one way i thought of, but I want it to work very fast, so the analog mult idea may not be capable of the speed. (the accuracy is not that important- 5% will do)

A mosfet has a square law relationship between voltage and current. You should be able to turn 6 dB/octave in input voltgage to 3db/octave in output current pretty easily. You can use it directly or in the feedback of an op amp. --DHeyward 21:34, 27 October 2007 (UTC)[reply]

This idea of using a mosfet (insted of an anlog mult) as the squaring element in MrR's topology is interesting. How would I setup a mosfet circuit to give me a true square of the input voltage (including the dc level)? Also could it be a JFET?

I just googled "mosfet square root circuit". here is one. The middle opamp has the mosfet in the feedback with the drain as the output. This is what I was thinking. JFETs are square law as well though they may require different biasing to stay in saturation. But now that I think about it, you may need square function to get you to "power" and then "power" is 3 db/Octave. --DHeyward 02:22, 28 October 2007 (UTC)[reply]

Interesting circuit, tho Im not keen on the necessary inversion and the active offsetting circuit. But thanks for finding it!

Doesn't "3dB per octave" just mean "a factor of 2 for each factor of 2"? I wonder why it isn't stated more simply as "1/f" or the like. —Tamfang 22:19, 29 October 2007 (UTC)[reply]

Whether that definition is right depends on what you mean by "a factor of 2". For a resistive load, the power is proportional to the square of the voltage. The OP wanted the output/input power ratio to change by a factor of 2 for each factor of 2 in frequency, which means the output/input voltage ratio is only changing by a factor of sqrt(2) for each factor of 2 in frequency. That's not as easy to do as to design a circuit such that the output/input voltage ratio changes by a factor of 2 for each factor of 2 in frequency. MrRedact 23:43, 29 October 2007 (UTC)[reply]

stop watch hardware

I want to know the harware of a stop watch(just stop watch) whose accuracy is 0.1 second. —Preceding unsigned comment added by Gangt (talk • contribs) 07:24, 27 October 2007 (UTC)[reply]

See stopwatch. The rate of a mechanical stopwatch is controlled by the balance wheel while an electronic stopwatch uses a crystal oscillator. --Justanother 19:14, 27 October 2007 (UTC)[reply]

Shock in AC and DC

Why do we get sucked in and pushed out while receiving shocks?..Is it due to ac and dc shock?..What is the shock difference in these two and which one is more fatal?.. —Preceding unsigned comment added by 122.164.49.85 (talk) 08:56, 27 October 2007 (UTC)[reply]

See Electric shock#Issues affecting lethality which discusses this. 84.64.123.72 11:33, 27 October 2007 (UTC)[reply]

The shock causes your muscles to contract. This can mean that your hand is immediately, and involuntarily pulled away from the cause of the shock - but if you have a grip around the live object then your grip may tighten and you'll be unable to let go no matter what. SteveBaker 02:44, 28 October 2007 (UTC)[reply]

German Weinstein: Potassium tartrate or Potassium bitartrate?

Hi, the German article de:Weinstein links to Potassium bitartrate aka cream of tartar etc. and vice versa. Dictionaries say so, too. (See also de:Wikipedia:Auskunft#Weinstein. But Potassium tartrate claims that that is the main ingredient of "Weinstein", though easily confused with the other. Can someone clear that up, please? T.a.k. 10:06, 27 October 2007 (UTC)[reply]

I'm not sure what is correct, but I would note that the information was added by an anonymous IP as that IP's only contribution [2]. Given that the information is not sourced, and contradicts other sources, I'm not sure that the info should be trusted. -- 16:45, 31 October 2007 (UTC)

Granular layer - cerebellum

Granular layer redirects to cerebellum. Why is this and is this correct? Lova Falk 10:32, 27 October 2007 (UTC)[reply]

Did you see the granular layer section? Here's a link with some other definitions. JMiall ₰ 11:48, 27 October 2007 (UTC)[reply]

Granular layers also occur outside of the cerebellum (most notably, in the cerebral cortex), so this redirection is not really appropriate I think. EelkeSpaak 15:32, 27 October 2007 (UTC)[reply]

Cost of heat

A man lives in a place where the temperature is below freezing all year round. His house is heated with electricity, and he has an electric water heater. He keeps his house at a constant 20C, except for the "cold room", in which the thermostat is set at 5C. The water heater is located in an area where the temperature is 20C. He decides to move the water heater to the "cold room", to gain more living space. (This does not significantly change the distance from the water heater to where the hot water is used.) The man notices that the temperature in the "cold room" remains at 5C, even though the water heater feels warm. Would the total monthly electric bill change because of the move? GarthGarth 13:31, 27 October 2007 (UTC)Garth[reply]

If I understand the assumptions correctly, no. Electric heaters are 100% efficient in the sense that they produce one kWh of heat per kWh of electricity. Since (by assumption) nothing else changed in the move, the total power consumed by the water and space heaters must be the same as before, although it may be differently distributed between them. -- BenRG 14:19, 27 October 2007 (UTC)[reply]

I can't see how that can be right. The hot water is loosing heat by means of radiation and conduction the rate of both are affected by the temperature difference between the hot water and the room temperature. Conduction is proportional to temperature difference and radiation goes as the 4th power.Theresa Knott | The otter sank 14:38, 27 October 2007 (UTC)[reply]

The water heater will have to work a little more to keep the water hot, compared to the 20C room, since more heat is being lost to the colder room. At the same time, the heater in the house has to work less to keep the 5C room at 5C, so the house heater will use less energy. So the monthly energy bill is probably going to change. Whether its up or down depends on more specific info such as efficiencies, volume/area of the rooms, etc, and a bit of calculus. Arakunem^Talk 15:26, 27 October 2007 (UTC)[reply]

The water heater will use more electricity due to the greater temperature differential between the hot water and the room. However, the wasted heat from the water heater will contribute to the heating of the room. So the room's heaters will work a little less, and the water heater will work a little more. Probably won't see much change in electrical usage.

But for the room the water heater was moved from, the room heater will have to work a little bit more than it did previously since the water heater is no longer contributing any heat. --Bowlhover 01:22, 30 October 2007 (UTC)[reply]

Obviously, this assumes the rooms are heated year round, and moving the water heater doesn't change the temperature of any of the rooms, and the new plumbing doesn't snake outside, and so on...--Duk 17:08, 27 October 2007 (UTC)[reply]

No, no change at all. Forget the details. The temperatures stay the same, so the same amount of heat is generated. The only problem might be efficiency, but since heat is the ultimate energy waste product, but also the goal here, efficiency will be 100%, no matter what. So unless more or less warmth leaks out of the house, for example because the heater is moved to or from an outer wall, energy consumption will remain the same. DirkvdM 17:52, 27 October 2007 (UTC)[reply]

That assumes that none of the heat from the water heater (and the water in it) is lost to the room. Arakunem^Talk 23:03, 27 October 2007 (UTC)[reply]

Actually, it's the opposite. It assumes the water heater is just another heating element in the room. There are 3. One for the 20C rooms, one for the 5C room and the water heater. The temperature doesn't change in the rooms. The heating elements were just rearranged. The details of the percentage of heat from each element is not asked, just the overall heat. It doesn't change because there are no changes in the temperatures of the rooms. The sum of the heat is the same in each case so the energy added is the same. --DHeyward 06:46, 28 October 2007 (UTC)[reply]

If the electricity is being turned to heat in both cases, there will be no difference. If the water heater works that way, but the heater from the room moves heat in from outside, which is more likely the case, the less efficient water heater will be doing proportionally more of the heating, and the electricity bill will go up. — Daniel 20:18, 27 October 2007 (UTC)[reply]

Hang on, hang on, hang on.

If the water heater loses some heat to the room, we can think of it as being less than 100% efficient at heating water, but it is guaranteed to be 100% efficient at heating the water+room system.

If the other furnace is outside, and if it is less than 100% efficient, it might well lose some heat to the outside, which would then be lost forever.

So, paradoxically, if the other heater is outside, the less efficient the water heater is, the lower the overall electricity bill will be...

(The only exception would be the case where the grossly inefficient water heater in the "cold room" ended up heating it above the requisite 5°C.)

But on the other hand, if the other heater is inside, and if it's less than 100% efficient, any heat it loses is lost to the room, so hey presto, it's 100% efficient after all. —Steve Summit (talk) 23:17, 27 October 2007 (UTC)[reply]

And then there's how leaky the windows might be... *shifty* —Preceding unsigned comment added by Arakunem (talk • contribs) 23:23, 27 October 2007 (UTC)[reply]

Makes no difference. There has to be some heat loss to the outside, otherwise waste heat from the water heater would heat the "cold room" above 5°, and for that matter the rest of the house above 20°. Whether that loss to the outside is slight or extreme is immaterial to the rest of the problem. —Steve Summit (talk) 01:03, 28 October 2007 (UTC)[reply]

...wait a minute. Did we just do someone's homework for him? —Steve Summit (talk) 01:04, 28 October 2007 (UTC)[reply]

It's okay. I think after reading this they'll be more confused, and have to resort to thinking through what their textbook says...Skittle 15:07, 28 October 2007 (UTC)[reply]

Ah - a homework question. Ok, just draw a box with a arrows for energy_in and energy_out. Since all the temperatures and house insulation stay the same, energy_out stays the same, and therefore electricity_in must stay the same. Forget everything else! --Duk 16:25, 28 October 2007 (UTC)[reply]

I don't think a water heater can ever be 100% efficient. It's heating water in copper tank which is connected via nice metal pipes off into the outside world. Being metal, those pipes are going to conduct heat away out of the tank - and out of the room too. The answer to this question depends on the relative efficiency of the room heater and the water heater. If the water heater is more efficient then your bills should go down - if it's less efficient then they'll go up. SteveBaker 02:42, 28 October 2007 (UTC)[reply]

Not so. See my "paradoxically" argument above. It'd be very hard for an electric water heater to lose heat directly to the outside world, without heating some part of the house up first. (But similarly for a resistive electric space heater, located inside the house.) —Steve Summit (talk) 03:44, 28 October 2007 (UTC)[reply]

Yes but thats the point. The heater has been moved to a colder room, therefore is losing more heat faster than it was in the warmer room. Thus, the heater's heater will be running more frequently to keep the water at its desired temp. Thus more power is used by the water heater simply because it is running more often. The second consideration is then, that the heat lost from the heater to the room means the room's heater will not need to work as hard to keep that room at 5C, so it will use less energy. So the total house energy bill will change, but without more info you cant say for sure which way it will change, and by how much. Arakunem^Talk 15:16, 28 October 2007 (UTC)[reply]

Explain to me again why it will cost more (or less) to heat a room with waste heat from an electric water heater, versus with heat from an electric heater. —Steve Summit (talk) 16:40, 28 October 2007 (UTC)[reply]

Yes, you are quite correct! I was looking at the components and not the system as a whole. *Awards one pie* :) Arakunem^Talk 18:21, 28 October 2007 (UTC)[reply]

I don't see the location making any difference in energy use in the specified, somewhat atypical, conditions. The 5 degree room is already electrically heated, so any heat lost from the hot water heater from the insulated tank or the water pipes or even the electrical wiring to the air in the cold room will displace heat which would otherwise have been consumed by the heater in the cold room which maintained it 5 degrees warmer than the outside. The heat from the water heater which would otherwise have helped heat the warm rooms will be replaced by additional heat from the electric heaters there. Zero net effect. If it were a real world home, and the water heater were moved from living space to an unheated garage or unheated basement, the electric bill might go up, since most homes do not have a room thermostatically maintained five degrees warmer than a constant temperature outside. The devil is in the details. Edison 21:10, 28 October 2007 (UTC)[reply]

Yeah, but the assumption was that the temperature in the rooms is thermostatically maintained.

I said it 'forget the details', Duk said that energy in = energy out and Arakunem said to look at the system as a whole. Same idea, the wording just improved gradually. If the energy in the system (the house) remains the same and the energy going out of the system (heat leakage from the house as a whole) remains the same, then the energy going in is the same. And all that energy would be in the form of heat, since heat is the ultimate energy waste product. It would have to be a very stupid heater if it didn't convert all the energy into heat - what else would it convert it into? DirkvdM 08:59, 29 October 2007 (UTC)[reply]

Light?

I have to disagree with you on this problem. When the water heater is moved from the hot room to the cold room, the house heater will be on more often in the original room since it must work independently to maintain the temperature. Conversely, the house heater will be on less often in the new room since the water heater is providing part of the heat. The total amount of electricity used by the house heater remains constant.

However, the water heater must keep its water at a constant temperature. More energy will be consumed if the air temperature is 5 degrees than if it is 20 degrees, and the newly-located heater will therefore use more energy than it did before. The conclusion is that more electricity will be used after the water heater is moved. Is my reasoning correct? --Bowlhover 01:22, 30 October 2007 (UTC)[reply]

No because the energy that is "lost" by the heater goes in to the room. Since the 5C room doesn't rise in temperature, the heat provided by the water heater offsets the heat that was used to heat the room to 5C. This is a rate problem where the sum of the heat entering the house equals the sum of the heat leaving the house. The temperature is constant so the heat added is constant. Nothing you do with location of the heat sources will change it. This is a Guassian sphere-like problem where the net heat flux is zero. --DHeyward 05:05, 30 October 2007 (UTC)[reply]

Yes, this is what got me earlier. Note that the heaters were specified to all be electric. Meaning X watts of electricity goes right into heat. If you look at the house as a system, rather than rooms, then the net energy change for the system will be zero. Look at it like this: If you had one big heater providing X amount of heat, that would be the same as having 2 smaller heaters providing X/2 heat, yes? Or 3 smaller heaters at X/3... The heat lost from the water heater is just another heat source. If the heater loses 100 watts to the room, thats 100 watts less that the room heater needs to furnish. Net change = zero. Arakunem^Talk 14:40, 30 October 2007 (UTC)[reply]

Why is hot water better for washing dishes?

Why is it that hot water is better when washing dishes? Why does it seem to remove more dirt/grease/debris than cold? Is it purely from the thermal effect? or does the sometimes higher pressure of a hot water tap jet also have an effect? 84.64.123.72 13:38, 27 October 2007 (UTC)[reply]

The main factor is that things dissolve more readily in hot water. For instance, a cup hot water can hold more sugar than a cup of cold water. This takes care of the dirt that dissolves in water. Most of the dirt (like fat) does not dissolve in water, so we add detergent to the water so these dissolve a well. Finally, some amount of mechanical force is necessary to speed the process along, which is where the brush comes in. This also means that if the hot water has greater pressure (not sure if that's true) it may have a slight effect. 14:14, 27 October 2007 (UTC) —Preceding unsigned comment added by Risk one (talk • contribs)

In addition hot water softens fat and allows emulsification with a soap or detergent more rapidly than at lower temeratures. Hey, it also makes doing the dishes more comfortable Richard Avery 15:13, 27 October 2007 (UTC)[reply]

"also makes doing the dishes more comfortable" - depends on how hot the hot water is. --Psud 00:53, 28 October 2007 (UTC)[reply]

Softening/melting fats is the primary reason. The liquid fat is emulsified by the detergent, this would take a lot long if the fat was still solid. Shniken1 17:19, 27 October 2007 (UTC)[reply]

Anyone else here use boiling water to clean out their frying pans? You don't even need soap if the water is hot enough - just pour it in from the kettle, let the fat melt and float to the top, pour it out, wipe the pan out with a cloth. --Kurt Shaped Box 18:54, 27 October 2007 (UTC)[reply]

Yep, also doing that avoids getting the pan too clean or something, according to my mother. And who am I to question her wisdom? Then you rub a tiny amount of oil into the pan's surface before storing it. And Dirk, I think you could only wash up with cold water if a) you didn't eat or cook any of the more interesting things I eat or b) didn't mind a thin film of rancid fat/grease on everything and c) didn't need to get things washed quickly and d) didn't have to concern yourself with germs. Skittle 15:02, 28 October 2007 (UTC)[reply]

Now that we've changed the subject into dishwashing tips, I could add one my husband sometimes applies: boil water and washing powder in a pan. The inside of the pan has never been so clean before! Lova Falk 19:06, 27 October 2007 (UTC)[reply]

A very slight effect, if any. Several years ago I started using cold water and I didn't notice a difference. Mind you, I do the dishes in two goes. I first apply the dishwater (with just a little washing up liquid, a fraction of what people in the US use), let that soak for 10 minutes while I do some other household task and then finish it off with ease. Mind you, I do this on the counter, not in the sink, with the dishwater in one of the pans, and then I dip the brush in that. Works perfectly. And then I have a good excuse to clean the counter when I'm ready. Win-win-win. :) DirkvdM 18:06, 27 October 2007 (UTC)[reply]

A side effect of using very hot water (at least for the final rinse) is that your dishes and glasses dry faster, with fewer water spots. —Steve Summit (talk) 18:30, 27 October 2007 (UTC)[reply]

If the water is sufficiently hot, it will also kill some bacteria. This may not be very relevant at home, but it becomes important if you're, say, running a summer camp where keeping the dishwashing water hot enough can mean the difference between one kid with diarrhea and 30 kids with diarrhea (and one outhouse). The problem is that the water needs to be actually uncomfortably, if not quite scaldingly, hot (at least around 50°C / 120°F or so; see e.g. [3], keeping in mind that chemical sanitizers are rarely available under camp conditions). Merely lukewarm water just makes the bugs grow faster. —Ilmari Karonen (talk) 23:27, 27 October 2007 (UTC)[reply]

You all seem to forget to rinse properly. Cold dishwater will just as easily dissolve all fat if given enough time. A brush (before and after the soak) will help this process. After that, rinse with a lot of water. Assuming any food stuff is soluble in either water or fat, this will remove everything and the pan should be as clean as the water. Right? Just try it with a very greasy pan. It would be nice if someone who has a petri dish at hand would try this. Btw, there is no need to remove/kill all bacteria. Actually, the rise of allergies in the last few decades has shown that it is actually a very bad idea. And in a normal life you're exposed to all sorts of sources of bacteria, so why become all clinical here? DirkvdM 09:09, 29 October 2007 (UTC)[reply]

electron transfer

what makes an electrin to be transfered from an atom to the other(e.g metal to non metal )in ionic bonding? —Preceding unsigned comment added by 195.225.63.210 (talk) 14:06, 27 October 2007 (UTC)[reply]

Systems of atoms like to be in the lowest possible energy state. When the electron moves from the metal to the non metal the total energy of the sytem goes down.

Another way of looking at it is thinking about the forces involved. Some atoms attract electrons more strongly than others. Metals tend to attract thier outer electrons less strongly than a non metal. See electronegativity for more detals. So although the electron is pupped towards the metal atom it is pulled more towards the non metal one and so hops over to it making an ionic bond Theresa Knott | The otter sank 14:29, 27 October 2007 (UTC)[reply]

After which they would each go on their own merry way, were it not that the metal is now positively charged and the non-metal positively charged, so they are attracted to each other. A bit like free sex - the exchange precedes the attraction. :) DirkvdM 18:12, 27 October 2007 (UTC)[reply]

plants

Why are some leaves spickey —Preceding unsigned comment added by 213.122.27.220 (talk) 18:22, 27 October 2007 (UTC)[reply]

A reason that comes quickly to mind is that the spiky leaves provide an adaptive advantage in protecting the plant against predation (being eaten) by animals. So, it basically acts as a self-defense mechanism in plants like holly for example. Azi Like a Fox 19:24, 27 October 2007 (UTC)[reply]

In other cases, and depending on how spiky you meant, water storage may also be a factor "Cacti have never lost their leaves completely; they have only reduced the size so that they reduce the surface area through which water can be lost by transpiration.". Our article on Leaves, specifically leaves#Adaptations, suggests that "A transformation into spines protects the plants" so my water storage comment may be a misnomer.--VectorPotential^Talk 12:33, 28 October 2007 (UTC)[reply]

What leaves are spiky? Be specific. Malamockq —Preceding comment was added at 17:17, 28 October 2007 (UTC)[reply]

The ventral posteromedial nucleus!

Here's another one of my detail questions!
On this picture the VPL borders the pulvinar, while the VPM is more in the middle. However, our own beloved Wikipedia has this picture, that puts the VPL in the middle and the VPM bordering the pulvinar. My guess is that the VPM should be in the middle (because it is called medial), and that the Wikipicture has got the VPM and the VPL mixed up.
Is my guess right or not? Lova Falk 18:53, 27 October 2007 (UTC)[reply]

Here is an actual brain section for Macaque showing the VPM medial to the VPL. This diagram also shows the VPM in a medial position. I don't think the Wikipedia diagram is meant to be very accurate....it smashes a 3D structure into a flat diagram. --JWSchmidt 04:36, 28 October 2007 (UTC)[reply]

Thank you! Lova Falk 13:48, 28 October 2007 (UTC)[reply]

Plant Absorbtion

I was wondering if plants could absorb anything besides water (ie milk, or perhaps coffee?). Would such liquids have a negative impact on plant growth or would their roots just absorb only the water part of the liquid? Thanks, Valens Impérial Császár 93 19:17, 27 October 2007 (UTC)[reply]

If a plant absorbed only water, it would die of malnutrition. Luckily for it, the water it's getting is generally dirty. Literally. I remember something about putting cellery in water dyed red with food coloring and watching the vain-like things that bring in the water turn red. — Daniel 19:51, 27 October 2007 (UTC)[reply]

You can do the same thing with flowers. Put a red rose's stem in water colored with blue ink, and you get a blue rose (according to my old biology textbook anyway). Most plants probably have a very narrow range of acidity that they can survive in. I know that mixing up used coffee grounds in soil for tomatoes helps them grow, but the caffeine in fresh coffee will kill them (so maybe decaf?). I expect that the fat content is important, which means that milk would be difficult. Most of this information (optimal acidity, hardness, etc) should be well documented for the soil, so it shouldn't be difficult to find a drink that suits a plant, or a plant that can grow in a given drink. risk 03:50, 28 October 2007 (UTC)[reply]

These are two different phenomena. Plants absorb the water and nutrients they need through their roots. When you cut the stem of a plant and immerse it in dye you are bypassing the roots, the liquid flows up through the plant by cappilary action. -- Diletante 20:48, 28 October 2007 (UTC)[reply]

How to kill Mad Cow prions?

I've heard that they're incredibly resiliant to heat, cold, UV light, pressure, chemicals and radiation. What's the best way of sterilizing something that's been contaminated with Mad Cow prions? —Preceding unsigned comment added by 84.71.69.70 (talk) 22:53, 27 October 2007 (UTC)[reply]

Our Prion#sterilization section suggets autoclaving at temperatures of 134 degrees Celsius for 18 minutes, which will pretty much denature any protein. If you want it utterly destroyed, incinerate it at even higher temperatures. If you're asking whether you can sterilize contaminated beef to an edible state, the answer is a flat no, as any method of destroying or deactivating the prion will destroy the beef. Someguy1221 22:59, 27 October 2007 (UTC)[reply]

A few years ago, I remember reading that heating contaminated medical instruments (to give an example) in concentrated sodium hydroxide or hydrofluoric acid at insanely high temperatures and pressures was the only sure-fire way of eradicating the infectious agents completely. I guess that technology has progressed somewhat since then. --Kurt Shaped Box 23:04, 27 October 2007 (UTC)[reply]

Minor point, but you can't kill a protein at all, it's not alive. If you denature a protein it can still, in most cases, be refolded under ideal conditions. See Protein denaturation, Protein folding, and of course Prion.--VectorPotential^Talk 12:26, 28 October 2007 (UTC)[reply]

Rather than temperature alone, please discuss the possibility of chemical neutralization of mad cow prions. Would bleach or diluted bleach neutralize prions? If someone fed the cat using one of the cereal bowls, and I want to make sure that prions which might have been in the "meat byproducts" part of the catfood do not remain on the dish when I eat my cereal from the same bowl sometime in the future, would a soak in bleach solution inactivate the prions? Would any other household chemical be more effective? Edison 21:01, 28 October 2007 (UTC)[reply]

You might find this paper and the references it contains useful. Of note:

"Conventional hospital disinfectants including ethylene oxide, propriolactone, hydrogen peroxide, iodophors, peracetic acid, chaotropes, and phenolics have little effect on prion infectivity.... In addition, prions are resistant to inactivation by UV irradiation, aldehyde fixation, boiling, standard gravity autoclaving at 121°C, and detergent solubilization."

As well,

"Currently recommended protocols for prion decontamination include either (i) >2% available chlorine of sodium hypochlorite for 2 h, (ii) 2 M NaOH for 1 h, or (iii) autoclaving at 134°C for 4.5 h. Each of these protocols has important limitations: sodium hypochlorite and NaOH are corrosive at the concentrations required to inactivate prions; NaOH did not inactivate CJD prions completely in some reports; and extended autoclaving at high temperature is deleterious to many materials. Currently, some high-risk surgical instruments are soaked in 2 N NaOH for 1 h, rinsed with water, and autoclaved at 134°C for 1 h, while many other such instruments are discarded."

The paper also presents a 'combined' protocol that employs an acidic SDS solution (1% sodium dodecyl sulfate plus 0.5% acetic acid) followed by 121°C that does seem to fully inactivate prions. Again, I urge you to read the full paper, and to look up the references it cites. TenOfAllTrades(talk) 21:26, 28 October 2007 (UTC)[reply]

As far as anyone knows, a human eating the meat will keep it from spreading. There's no proof humans can get infected from eating meat from a cow with mad cow disease. — Daniel 00:29, 29 October 2007 (UTC)[reply]

Just to inject a little realism....any idea how many people have contracted vCJD (human form of mad cow disease)? Only about 160, total, worldwide. Almost all of these lived in the UK during the mad cow epidemic. The number of cases annually, world wide, is quickly approaching zero. You know how many domesically born cattle have been found in the US with mad cow disease? 2!! [4]. In my opinion, this shouldn't be high on your list of worries. ike9898 21:12, 29 October 2007 (UTC)[reply]

When our research group needs to really and truly destroy protein residues we bake metal or glassware at 400 C for 2 hours in oxygen. This converts all the carbon to carbon dioxide. Of course, there are lots of other materials which also won't survive being steralized in this way, so its practical utility may be limited. Dragons flight 09:16, 30 October 2007 (UTC)[reply]

Dropping uncertainties

When dividing two numbers with uncertainties, the end result should have a greater uncertainty right? However, when I am doing the following:

[(2.91 ± 0.01g) / (2.55 ± 0.01g)]

I end up with:

1.14 ± 0.008g

which is lower than the two original uncertainties I had. Could this be correct?

Thanks. Acceptable 23:15, 27 October 2007 (UTC)[reply]

When dividing two numbers with uncertainties, the end result should have a greater relative uncertainty, which your answer indeed does. However, I get 1.141 ± 0.006 for the answer. Did you make an arithmetic error? —Keenan Pepper 23:30, 27 October 2007 (UTC)[reply]

Ah yes, a greater relative uncertainty makes more sense. I did the calculation several times and I'm almost certain that 0.008 is correct. Much thanks. Acceptable 23:42, 27 October 2007 (UTC)[reply]

I actually get something a bit higher than 0.008:

Max value = 2.92/2.54 = 1.1496

Min value = 2.90/2.56 = 1.1328

Median value = (1.1496 + 1.1328)/2 = 1.1412 ± .0084. StuRat 18:50, 28 October 2007 (UTC)[reply]

But unfortunately Stu that is not how errors are calculated, although it seems logical. The errors in science are not upper and lower absolute bounds but probabilities basd on assumed normal distributions. Our page on error propagation gives the general formula for calculating errors. Under the specific examples you can see that for a ratio, the fractional errors combine in quadrature. I get 1.1412 +- 0.0068. Cyta 21:16, 28 October 2007 (UTC)[reply]

Which, if rounded to 2 decimal places (the error, that is), would give 1.1412 ± .007. -- JackofOz 00:14, 29 October 2007 (UTC)[reply]

How did you get 0.0068? The relative uncertainty of the numerator is 0.01 / 2.91 = 0.0034. The relative uncertainty of the denominator is 0.01 / 2.55 = 0.0039. So the relative uncertainty of the ratio should be sqrt(0.0034^2 + 0.0039^2) = 0.0052. The ratio is 1.141, so the absolute uncertainty should be 0.0052 * 1.141 = 0.0059, yielding 1.141 +/- 0.006. Right? —Keenan Pepper 22:06, 29 October 2007 (UTC)[reply]

the problem here is that the conventional nomenclature fro uncertainty (x±y) appears to be unambiguous, but actually is not.

You must explicitly define what you mean when you use the "x±y" Nomenclature. For example, if "(x±y)" means "x, with an assumption of a normal distribution with an SD of y" then "(x±y)/(z±w)" has a defined meaning. If "(x±y)" has a different meaning, then "(x±y)/(z±w)" has a different meaning. Note that you need to understand both the statistics of the numerator and the statistics of the denominator before you can have any confidence in the statistics of the result.-Arch dude 03:21, 29 October 2007 (UTC)[reply]

Excellent point, Arch dude. However, in the absence of any other statistical information it's reasonable to assume the variables are independent and normally distributed (or log normal, which is practically equivalent if the standard deviation is small compared to the value, as in this case). So adding the relative errors in quadrature is a good rule of thumb. —Keenan Pepper 22:06, 29 October 2007 (UTC)[reply]